Question

Using a suitable identity, factorise :
p (q=r)+q (r-p)2 + p (p-q)»

Answer 1

Answer:

Solution :

p³ (q - r)³ + q³ (r - p)³+r³(p - q)³

⇒ [p (q - r)]³+[q (r - p)]³+ [r (p - q)]³

__

if,

a = p (q - r) , b = q (r - p) , c = r (p - q)

By adding a , b & c,

We get,

⇒ a + b + c = p (q - r) + q (r - p) + (p - q) = (pq - pr) + (qr - pq) + (pr - qr) = pq - pq + qr - qr + pr - pr = 0

We know that,

If a + b + c = 0,

then,

a³ + b³ + c³ = 3abc (Identity)

So,

By substituting,

a = p (q - r) , b = q (r - p) , c = r (p - q)

We get,

⇒ [p(q - r)]³ + [q(r - q)]³ + [r(p - q)]³ = 3pqr (q - r)(r - p)(p - q)

⇒p³ (q - r)³ + q³ (r - p)³ + r³ (p - q)³ = 3pqr (q - r)(r - p)(p - q)

The factorized form of p³ (q - r)³ + q³ (r - p)³ + r³ (p - q)³ is 3pqr (q - r)(r - p)(p - q)